Add.
-$$\frac { 2 } { x ^ { 2 } - 3 x + 2 } + \frac { 7 } { x ^ { 2 } - 1 }$$
A) $\frac { 12 x - 9 } { ( x - 1 ) ( x + 1 ) ( x - 2 ) }$
B) $\frac { 28 x - 12 } { ( x - 1 ) ( x + 1 ) ( x - 2 ) }$
C) $\frac { 9 x - 12 } { ( x - 1 ) ( x - 2 ) }$
D) $\frac { 9 x - 12 } { ( x - 1 ) ( x + 1 ) ( x - 2 ) }$

Question

Quizplus · Accepted Answer

We will factor both denominators:

$$\frac { 2 } { x ^ { 2 } - 3 x + 2 } + \frac { 7 } { x ^ { 2 } - 1 } = \frac{2}{(x-1)(x-2)} + \frac{7}{(x-1)(x+1)}$$

Now, we need to find a common denominator for these two fractions. The easiest way to do this is to take the product of the two denominators:

$$(x-1)(x-2)(x+1)$$

We can then rewrite the original expression with this common denominator:

$$\frac{2(x+1)}{(x-1)(x-2)(x+1)} + \frac{7(x-2)}{(x-1)(x+1)(x-2)}$$

Now we can add the two fractions:

$$\frac{2(x+1)}{(x-1)(x-2)(x+1)} + \frac{7(x-2)}{(x-1)(x+1)(x-2)} = \frac{2(x+1)(x+1)}{(x-1)(x-2)(x+1)(x+1)} + \frac{7(x-2)(x-1)}{(x-1)(x+1)(x-2)(x-1)}$$
$$\frac{2(x^2+2x+1)}{(x-1)(x-2)(x+1)(x+1)} + \frac{7(x^2-3x+2)}{(x-1)(x+1)(x-2)(x-1)}$$
$$\frac{2x^2+4x+2}{(x-1)(x-2)(x+1)(x+1)} + \frac{7x^2-21x+14}{(x-1)(x+1)(x-2)(x-1)}$$

Now, we can combine the two fractions into one by finding a common denominator in the usual way:

$$\frac{(2x^2+4x+2)(x+1)(x-1)(x-2) + (7x^2-21x+14)(x+1)(x-1)(x+1)}{(x-1)(x+1)(x-2)(x-1)(x+1)(x+1)}$$

Simplifying the numerator and canceling out common factors, we get:

$$\frac{9x-12}{(x-1)(x+1)(x-2)}$$

Therefore, the correct answer is D, $\frac{9x-12}{(x-1)(x+1)(x-2)}$.

Add - 2x2−3x+2+7x2−1\frac { 2 } { x ^ { 2 } - 3 x + 2 } + \frac { 7 } { x ^ { 2 } - 1 }x2−3x+22​+x2−17​

Add
- $\frac { 2 } { x ^ { 2 } - 3 x + 2 } + \frac { 7 } { x ^ { 2 } - 1 }$